高质量题解|A30386.菲波那契数列
2026-01-28 11:35:22
发布于：北京
解题思路
斐波那契数列就是当前数字是前1个数和前2个数的和
所以我们的返回值就是f(n - 2) + f(n - 1)
斐波那契数列是从第3项开始的，所以边界条件就是n == 1 || n == 2
if(n == 1 || n == 2){
return 1;
}
return f(n - 2) + f(n - 1);
这就是函数内要写的
代码
#include <bits/stdc++.h>
using namespace std;
int f(long long n){
if(n == 1 || n == 2){
return 1;
}
return f(n - 2) + f(n - 1);
}
int main(){
long long n;
cin >> n;
cout << f(n);
return 0;
}
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